Jar A has x litres milk and jar B has y litres water. `80%` milk and `20%`water was taken out from the respective jars and were mixed in jar C. The respective ratio between milk and water in jar C was `2:1`. When 28 litres pure milk was added to jar C, the total quantity of mixture in jar C became 76 litres. What was the value of x?

Options

A. 50
B. 60
C. 80
D. 40

Answer

To solve the problem step by step, we will break down the information given and derive equations based on the conditions provided.

### Step 1: Define the quantities in jars
Let:
- Jar A has \( x \) litres of milk.
- Jar B has \( y \) litres of water.

### Step 2: Calculate the quantities taken out from jars
From Jar A, 80% of the milk is taken out:
- Milk taken from Jar A = \( 0.8x \)

From Jar B, 20% of the water is taken out:
- Water taken from Jar B = \( 0.2y \)

### Step 3: Determine the quantities in Jar C
The quantities mixed in Jar C are:
- Milk in Jar C = \( 0.8x \)
- Water in Jar C = \( 0.2y \)

### Step 4: Set up the ratio of milk to water in Jar C
According to the problem, the ratio of milk to water in Jar C is \( 2:1 \):
\[ \frac{0.8x}{0.2y} = \frac{2}{1} \]

### Step 5: Cross-multiply to find a relationship between \( x \) and \( y \)
Cross-multiplying gives:
\[ 0.8x = 0.2y \cdot 2 \]
\[ 0.8x = 0.4y \]
Dividing both sides by 0.4:
\[ 2x = y \quad \text{(Equation 1)} \]

### Step 6: Analyze the addition of milk to Jar C
When 28 litres of pure milk is added to Jar C, the total mixture becomes 76 litres:
\[ 0.8x + 28 + 0.2y = 76 \]

### Step 7: Substitute \( y \) from Equation 1 into the mixture equation
Substituting \( y = 2x \) into the equation:
\[ 0.8x + 28 + 0.2(2x) = 76 \]
\[ 0.8x + 28 + 0.4x = 76 \]
Combining like terms:
\[ 1.2x + 28 = 76 \]

### Step 8: Solve for \( x \)
Subtract 28 from both sides:
\[ 1.2x = 76 - 28 \]
\[ 1.2x = 48 \]
Now, divide both sides by 1.2:
\[ x = \frac{48}{1.2} = 40 \]

### Conclusion
The value of \( x \) is \( 40 \) litres.

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